Chapter 3.3, Problem 3.85P

A worker tries to move a rock by applying a 360-N force to a steel bar as shown, (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of Part a.

To determine

An equivalent force-couple system applied at D to replace the force at B.

Answer to Problem 3.85P

Thus the force-couple system at D to replace the force at B is $-231.40\text{N}\widehat{\text{i}}-275.78\text{N}\widehat{\text{j}}$, $230.45\text{N}\cdot \text{m}\widehat{\text{k}}$.

Explanation of Solution

Refer Fig P3.85.

Write the expression of the force at D in component form.

${\overrightarrow{F}}_B=F_B\left(-\sin \theta \widehat{\text{i}}-\cos \theta \widehat{\text{j}}\right)$ (1)

Here $F_B$ is the magnitude of the force applied at B, $\theta$ is the angle made by the force with the vertical, ${\overrightarrow{F}}_B$ is the force at B.

Write an expression of the vector connecting point B and D.

$\overrightarrow{r}=d_{\text{BD}}\left(-\cos \alpha \widehat{\text{i}}\text{+sin}\alpha \widehat{\text{j}}\right)$ (2)

Here $\overrightarrow{r}$ is the vector joining point B and D, $d_{\text{BD}}$ is the distance between point B and D, $\alpha$ is the angle made by the steel bar with the horizontal.

Write an expression for the couple at D.

$\overrightarrow{M}=\overrightarrow{r}\times \overrightarrow{F}$ (3)

Here $\overrightarrow{M}$ is the couple at D, $\overrightarrow{r}$ is the vector joining point B and D.

Conclusion:

Substitute $360\text{N}$ for $F_B$, $40°$ for $\theta$ to determine ${\overrightarrow{F}}_B$ in equation (1).

$\begin{array}{c}{\overrightarrow{F}}_B=360\text{N}\left(-\sin 40°\widehat{\text{i}}-\cos 40°\widehat{\text{j}}\right)\\ =-231.40\text{N}\widehat{\text{i}}-275.78\text{N}\widehat{\text{j}}\end{array}$

Substitute $0.65\text{m}$ for $d_{\text{BD}}$, $30°$ for $\alpha$ in equation (2)

$\begin{array}{c}\overrightarrow{r}=\left(0.65\text{m}\right)\left(-\cos 30°\widehat{\text{i}}\text{+sin}30°\widehat{\text{j}}\right)\\ =-0.56292\text{m}\widehat{\text{i}}\text{+0}\text{.32500}\text{m}\widehat{\text{j}}\end{array}$

Substitute the value of ${\overrightarrow{F}}_B$ and $\overrightarrow{r}$ for $\overrightarrow{M}$.

$\begin{array}{c}\overrightarrow{M}=\left|\begin{array}{ccc}\widehat{\text{i}}& \widehat{\text{j}}& \widehat{\text{k}}\\ -0.56292\text{m}& \text{0}\text{.32500}\text{m}& 0\\ -231.40\text{N}& -275.78\text{N}& 0\end{array}\right|\\ =230.45\text{N}\cdot \text{m}\widehat{\text{k}}\end{array}$

Thus the force-couple system at D to replace the force at B is $-231.40\text{N}\widehat{\text{i}}-275.78\text{N}\widehat{\text{j}}$, $230.45\text{N}\cdot \text{m}\widehat{\text{k}}$.

To determine

Two forces at A and D to replace the force in section (a).

Answer to Problem 3.85P

Thus the vertical force at A is $253.42\text{N}\widehat{\text{j}}$, and at D is $232.48\text{N}$ making an angle of $5.5193°$ with the x-axis.

Explanation of Solution

Refer Fig P3.85.

Write the expression of couple at D.

${\overrightarrow{M}}_D={\overrightarrow{r}}_{AD}\times {\overrightarrow{F}}_A$ (1)

Here ${\overrightarrow{M}}_D$ is the couple at D, ${\overrightarrow{r}}_{AD}$ is the vector joining A and D, ${\overrightarrow{F}}_A$ is the force at A.

Write the expression of the ${\overrightarrow{r}}_{AD}$.

${\overrightarrow{r}}_{AD}=-\left(d_{AD}\right)\left(\cos \alpha \widehat{\text{i}}\text{+sin}\alpha \widehat{\text{j}}\right)$ (2)

Here $d_{AD}$ is the distance between A and D, $\alpha$ is the angle between the steel bar with the horizontal.

Write the expression connecting the forces at A, D and B with the force in the section (a).

$\overrightarrow{F}={\overrightarrow{F}}_A+{\overrightarrow{F}}_D$ (3)

Here $\overrightarrow{F}$ is the force in the section (a), ${\overrightarrow{F}}_A$ is the force at A, ${\overrightarrow{F}}_D$ is the force at D.

Conclusion:

Substitute $1.05\text{m}$ for $d_{AD}$, $30°$ for $\alpha$ in equation (2).

$\begin{array}{c}{\overrightarrow{r}}_{AD}=-\left(1.05\text{m}\right)\left(\cos 30°\widehat{\text{i}}\text{+sin}30°\widehat{\text{j}}\right)\\ =-0.90933\text{m}\widehat{\text{i}}\text{+0}\text{.52500}\text{m}\widehat{\text{j}}\end{array}$

Substitute $230.45\text{N}\cdot \text{m}\widehat{\text{k}}$ for ${\overrightarrow{M}}_D$, $-0.90933\text{m}\widehat{\text{i}}\text{+0}\text{.52500}\text{m}\widehat{\text{j}}$ for ${\overrightarrow{r}}_{AD}$ in equation (1).

$\begin{array}{c}230.45\text{N}\cdot \text{m}\widehat{\text{k}}=\left|\begin{array}{ccc}\widehat{\text{i}}& \widehat{\text{j}}& \widehat{\text{k}}\\ -0.90933\text{m}& \text{0}\text{.52500}\text{m}& 0\\ 0& -F_A& 0\end{array}\right|\\ =0.90933F_A\text{m}\widehat{\text{k}}\end{array}$

Solve the above equation to get $F_A$.

$\begin{array}{c}{F}_A=\frac{230.45\text{N}\cdot \text{m}\widehat{\text{k}}}{0.90933\text{m}\widehat{\text{k}}}\\ =253.42\text{N}\end{array}$

Substitute $253.42\text{N}\widehat{\text{j}}$ for ${\overrightarrow{F}}_A$, $-231.40\text{N}\widehat{\text{i}}-275.78\text{N}\widehat{\text{j}}$ for $\overrightarrow{F}$ to determine ${\overrightarrow{F}}_D$.

$\begin{array}{c}{\overrightarrow{F}}_D=-231.40\text{N}\widehat{\text{i}}-275.78\text{N}\widehat{\text{j}}\text{+}253.42\text{N}\widehat{\text{j}}\\ =-231.40\text{N}\widehat{\text{i}}-22.36\text{N}\widehat{\text{j}}\end{array}$

Write the expression for the magnitude of ${\overrightarrow{F}}_D$.

$\begin{array}{c}{F}_D=\sqrt{{\left(-231.40\text{N}\right)}^2+{\left(-22.36\text{N}\right)}^2}\\ =232.48\text{N}\end{array}$

Write the expression for the angle made by ${\overrightarrow{F}}_D$ with x-axis.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{22.36\text{N}}{231.40\text{N}}\right)\\ =5.5193°\end{array}$

Thus the vertical force at A is $253.42\text{N}\widehat{\text{j}}$, and at D is $232.48\text{N}$ making an angle of $5.5193°$ with the x-axis.